At what distance from a carbon nucleus is the electric potential 1.8 V? Carbon atoms have 6 electrons in the atomic shell
...im very confused with this question, please help!
I have consulted with to professional PhD physicists on this, they and I are of the opinion your teacher as asked a question that has no easy answer. You have to have a model for the electron cloud around the nucleus, and with a probability distribution, this is going to be NOT introductory physics modeling or analysis. Perhaps your teacher had something in mind other than what was really asked. Sorry.
u use the equation V=kq/r......where k is 8.99*10^9 and q=1.6*10^-19 but remember to multiply q by 6 and solve for r
To determine the distance from a carbon nucleus at which the electric potential is 1.8 V, we need to consider the fact that the electric potential decreases with distance from the nucleus.
The electric potential, V, due to a point charge, can be calculated using the formula:
V = k * (Q / r)
Where:
- V is the electric potential
- k is the Coulomb's constant (approximately 8.99 x 10^9 Nm^2/C^2)
- Q is the charge of the nucleus (in this case, taken as +6e, where e is the elementary charge)
- r is the distance from the nucleus
Since the carbon atom has 6 electrons in its atomic shell, the nucleus contains 6 protons, each with a charge of +e. Thus, the charge Q of the nucleus is 6 times the elementary charge.
Now, let's rearrange the formula to solve for r:
r = k * (Q / V)
Substituting the given values:
Q = 6e
V = 1.8 V
k = 8.99 x 10^9 Nm^2/C^2
r = (8.99 x 10^9 Nm^2/C^2) * (6e / 1.8 V)
To simplify the calculation, we need to convert the elementary charge (e) to coulombs:
1 e = 1.6 x 10^(-19) C
r = (8.99 x 10^9 Nm^2/C^2) * (6 * 1.6 x 10^(-19) C) / 1.8 V
Calculating this expression will give us the distance (r) in meters from the carbon nucleus at a potential of 1.8 V.